We are asked to evaluate
the line integral along the curve C
of F dot differential r where the vector field F
is given here and the curve C is the circle
at radius four centered at the origin
parametrized in the counterclockwise direction
as shown here.
One way to interpret the meaning
of the value of a line integral
would be the amount of work the force field
or vector field does on a particle
as it travels around the circle with this orientation.
Looking at this graphically,
the vector field F is graphed in purple,
and the curve C is graphed in red.
If we take a look at this point here,
the value of the line integral represents
the amount of work the vector field
or force field does on the particle
as it travels around the circle.
Going back to our work, the first step
is to find the vector function r of t
that will trace out the curve C in the XY plane.
Remember, the parametric equations
for a circle centered at the origin
are X equals r cosine t and Y equals r sine t.
Or in our case, r is equal to four
which means r of t has an x component of four cosine t
and a y component of four sine t.
And to have a counterclockwise orientation,
we would have t greater than or equal to zero
and less than or equal to two pi radians.
So looking at our notes below,
the line integral along the curve C
of F dot differential r equals the integral
from a to b of F of x of t comma y of t
dotted with r prime of t differential t.
Which means the next step is to write
the vector field F as a function of t
using x equals four cosine t and y equals four sine t.
The given x component is three x minus y
which means in terms of t we have
three times four cosine t minus four sine t.
And therefore, we have an x component
of 12 cosine t
minus four sine t.
The given y component is x plus five y.
In terms of t, we have four cosine t
plus five times four sine t
which gives us a y component of four cosine t
plus 20 sine t.
We also need to find r prime of t.
Let's go ahead and do that now.
The x component is equal to the derivative
of four cosine t which is negative four sine t.
And the y component is equal to the derivative
of four sine t which is four cosine t.
And now we have all the information that we need.
The given line integral is equal
to the integral from a to b which is
from zero to two pi
of F of x of t comma y of t
dotted with r prime of t.
And F of x of t comma y of t has an x component
of 12 cosine t
minus four sine t
and a y component of four cosine t
plus 20 sine t.
We're going to dot this with r prime of t
which has an x component of negative four sine t
and a y component of four cosine t.
Now let's go to the next slide
and determine this dot product.
Determining the dot product, we have the quantity
12 times cosine t minus four sine t
times negative four sine t.
And then, plus the quantity four cosine t
plus 20 sine t
times four cosine t.
If we want to do distribution, we have
negative 48 cosine t sine t,
and then plus 16 sine-squared t.
And here we distribute four cosine t
which gives us plus 16 cosine-squared t
and then plus 80 cosine t sine t.
Focusing on this sum here, we can factor out 16
which gives us 16 times the quantity
sine-squared t plus cosine-squared t.
We know sine-squared t plus cosine-squared t
is equal to one, and therefore,
this sum simplifies to 16.
And then we have negative 48 cosine t sine t
plus 80 cosine t sine t
which is equal to 32 cosine t sine t.
All of this simplifies to 16 plus 32 cosine t sine t.
So going back to the first slide,
we now know this integral is equal to
the integral from zero to two pi
of 16 plus 32 cosine t sine t
differential t.
Let's evaluate this on the next slide.
To integrate the second term,
we'll need to perform u substitution.
So let's break this up into two separate integrals.
We have the integral from zero to two pi of 16
plus the integral from zero to two pi of 32 cosine t sine t.
The antiderivative of 16 with respect to t is 16 t.
To integrate here, we need to perform u substitution.
Let's let u equal sine t.
And therefore, differential u is equal to cosine t dt.
So all of this is just 32,
and cosine t dt is du, and this is u.
So we have 32 u differential u.
The antiderivative of 32 u with respect to u
is 32 times u-squared divided by two
which simplifies to 16 u-squared.
So we have plus 16 u-squared,
but u-squared is really sine-squared t.
And now we can find big F of b
minus big F of a.
Here, we factor out the 16,
we have 16 times the quantity two pi minus zero.
Here if we factor out the 16,
we have plus 16 times the quantity
sine-squared two pi minus sine-squared zero.
Well, sine two pi is zero and so is sine zero.
So all this simplifies to zero.
And this simplifies nice here
to just 16 times two pi which is equal to 32 pi.
Going back to the graph one last time,
we can think of this value
as the amount of work that the purple vector field
does on this blue particle as it travels
around the red circle.
I hope you found this helpful.